Answer to Question #347514 in Statistics and Probability for Yango

Question

A bank Manager wishes to provide a prompt service to customers at the bank derive up windows. The bank currently can serve up to 10 customers per 15 minutes period without significant delay. The average arrival rate is 7 customers per 15 minutes period. Let 𝑥 be the number of customers arriving per 15 minutes period. Assuming that 𝑥 has a Poisson distribution. a) Find

i. The probability that 10 customers will arrive in a particular 15 minutes period. ii. The probability that 10 or fewer customers will arrive in a particular 15 minutes period.

b) Find the probability that there will be a significantly delay at the derive up window. That is to find the probability that more than 10 customers will arrive during a particular 15 minutes period.

Answer 1

a) $\lambda t=7$

i.

$$P(X=10)=\dfrac{e^{-7}(7)^{10}}{10!}=0.07098$$

ii.

$$P(X\le10)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)+P(X=4)+P(X=5)$$

$$+P(X=6)+P(X=7)+P(X=8)$$

$$+P(X=9)+P(X=10)$$

$$=\dfrac{e^{-7}(7)^0}{0!}+\dfrac{e^{-7}(7)^1}{1!}+\dfrac{e^{-7}(7)^2}{2!}$$

$$+\dfrac{e^{-7}(7)^3}{3!}+\dfrac{e^{-7}(7)^4}{4!}+\dfrac{e^{-7}(7)^5}{5!}$$

$$+\dfrac{e^{-7}(7)^6}{6!}+\dfrac{e^{-7}(7)^7}{7!}+\dfrac{e^{-7}(7)^8}{8!}$$

$$+\dfrac{e^{-7}(7)^9}{9!}+\dfrac{e^{-7}(7)^{10}}{10!}=0.90148$$

b)

$$P(X>10)=1-P(X\le 10)$$

$$=0.09852$$